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Question: Let $A$ be a $3*2$ matrix with $rank(A)=2$. Let $\sigma_1$ and $\sigma_2$ denote the singular values of $A$. Let {$\vec{v_1},\vec{v_2}$} be an orthonormal basis for $R^2$ of right singular vectors of $A$. Let $\vec{u_1}=\frac{1}{\sigma_1}A\vec{v_1}$ and $\vec{u_2}=\frac{1}{\sigma_2}A\vec{v_2}$.

Prove that if $\vec{u_3}=\vec{u_1}\times\vec{u_2}$ then $\vec{u_3}$ is a left singular vector of A.

My solution: Can't even get started on it. The only thing I can think of is that both the singular values should not be 0 as they appear in the denominator. This means we only have to use one definition of singular matrix.

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Recall that a right singular vector is an eigenvector of $A^*A$ and a left singular vector is an eigenvector of $AA^*$. Then observe $$ AA^* u_i = \frac1{\sigma_i} AA^*A v_i = Av_i = \sigma_i u_i$$ So $u_{1},u_2$ are left singular vectors that are linearly independent. As the left singular vectors are orthogonal, the cross product is the third and final left singular vector.

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  • $\begingroup$ In the first equation how does $\frac{1}{\sigma_i}AA^TA\vec{v_i}=A\vec{v_i}$? $\endgroup$ – kronos Jul 12 '18 at 1:47
  • $\begingroup$ @kronos $v_i$ is an eigenvector of $A^TA$ $\endgroup$ – Calvin Khor Jul 12 '18 at 6:09

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